# Design of Control Systems

 Question 1
The transfer function of a phase lead compensator is given by
$D(s)=\frac{3\left ( s+\frac{1}{3T} \right )}{\left ( s+\frac{1}{T} \right )}$
The frequency (in rad/sec), at which $\angle D(j\omega )$ is maximum, is
 A $\sqrt{\frac{3}{T^2}}$ B $\sqrt{\frac{1}{3T^2}}$ C $\sqrt{3T}$ D $\sqrt{3T^3}$
GATE EE 2019   Control Systems
Question 1 Explanation:
$T(s)=\frac{1+3Ts}{1+Ts}$
Frequency at which $\angle T(j\omega )$ is maximum,
$(\omega _m)=\frac{1}{T\sqrt{\alpha }}$
$\alpha =\frac{1}{1/3}=3$
$\omega _m=\frac{1}{T\sqrt{3}}=\sqrt{\frac{1}{3T^2}}$
 Question 2
The transfer function C(s) of a compensator is given below.
$C(s)=\frac{(1+\frac{s}{0.1})(1+\frac{s}{100})}{(1+s)(1+\frac{s}{10})}$
The frequency range in which the phase (lead) introduced by the compensator reaches the maximum is
 A $0.1 \lt \omega \lt 1$ B $1 \lt \omega \lt 10$ C $10 \lt \omega \lt 100$ D $\omega \gt 100$
GATE EE 2017-SET-2   Control Systems
Question 2 Explanation:
Pole zero diagram of compensator transfer function is shown below.

Maximum phase lead is between 0.1 and 1.
$0.1 \lt \omega \lt 1$

 Question 3
For the network shown in the figure below, the frequency (in rad/s) at which the maximum phase lag occurs is, ___________.
 A 0.15 B 0.32 C 0.66 D 0.92
GATE EE 2016-SET-2   Control Systems
Question 3 Explanation:
Assuming
\begin{aligned} R_1T&=9\Omega \\ R_2&=1\Omega\\ \text{ We can write,} \\ \frac{V_o(s)}{V_{in}(s)}&=\frac{R_2+\frac{1}{sC}}{R_1+R_2+\frac{1}{sC}} \\ &=\frac{1+R_2\cdot Cs}{1+(R_1+R_2)Cs} \\ &=\frac{1+R_2\cdot Cs}{1+\left (\frac{R_1+R_2}{R_2} \right )R_2 Cs}\\ \text{Let, } R_2C&=T \\ \frac{R_1+R_2}{R_2}&=\beta \\ \text{Hence, } \\ \frac{V_o(s)}{V_{in}(s)}&=\frac{1+Ts}{(1+\beta Ts)} \\ \text{Which represent }& \text{a lag compensator}\\ \because \text{ Here, } T&=R_2C=1.1=1 sec \\ \beta &=\frac{1+9}{1}=10\\ \text{Maximum phase }& \text{lag occurs at frequency}\\ \omega _n&=\frac{1}{T\sqrt{\beta }} \\ &=\frac{1}{1\sqrt{10}} =0.316 rad/sec. \end{aligned}
 Question 4
The transfer function of a compensator is given as
$G_c(s)=\frac{s+a}{s+b}$
The phase of the above lead compensator is maximum at
 A $\sqrt{2}$ rad/s B $\sqrt{3}$ rad/s C $\sqrt{6}$ rad/s D $1/ \sqrt{3}$ rad/s
GATE EE 2012   Control Systems
Question 4 Explanation:
$\omega _m=\sqrt{\omega _{c1} \times \omega _{c2}}$
$\;\;=\sqrt{1 times 2}=\sqrt{2}$ rad/sec.
 Question 5
The transfer function of a compensator is given as
$G_c(s)=\frac{s+a}{s+b}$
$G_c(s)$ is a lead compensator if
 A a = 1, b = 2 B a = 3, b = 2 C a =- 3, b =- 1 D a = 3, b = 1
GATE EE 2012   Control Systems
Question 5 Explanation:
$G_c(s)=\frac{s+a}{s+b}$
$G_c(j\omega )=\frac{j\omega +a}{j\omega +b}$
$\angle G_c(j\omega )=tan^{-1}\frac{\omega }{a}-tan^{-1}\frac{\omega }{b}$
$\;\;\;= tan^{-1}\left ( \frac{\frac{\omega }{a}-\frac{\omega }{b}}{1+\frac{\omega ^2 }{ab}} \right )$
For $G_c(s)$ to be a lead compensator
$\angle G_c(j\omega ) \gt 0$
$\frac{\omega }{a} \gt \frac{\omega }{b}$
$\Rightarrow b \gt a$
Option (A) Satisfies the above equation.

There are 5 questions to complete.